Posted by **Priscilla** on Friday, December 25, 2009 at 10:43pm.

A race car can be slowed with a constant acceleration of -11 m/s^2

a. If the car is going 55 m/s, how many meters will it travel before it stops?

b. How many meters will it take to stop a car going twice as fast?

- Physics -
**Damon**, Saturday, December 26, 2009 at 4:18am
v = Vo + a t

so

v = 55 - 11 t

when is v = 0 ?

0 = 55 - 11 t

so t = 5 seconds to stop

d = Vo t + (1/2) a t^2

d = 55 (5) - (1/2) 11 (25)

= 275 - 137.5

=137.5 meters

Now do that again for Vo = 110 m/s

- Physics -
**Heart**, Saturday, December 26, 2009 at 8:49am
Ok, thanks but i don't understand b. I know you use the equation d = Vit +1/2 at^2 but to find the distance would you just calculate d = vit only.

- Physics -
**Damon**, Saturday, December 26, 2009 at 2:03pm
No, do v = Vo + a t again with

0 = 110 - 11 t

then use that t in

d = 110 t - (1/2)11 t^2

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